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<p>How to compute the Fourier coefficients? Use the orthogonality of the trig set!<dfn class="terminology">Definition</dfn> Given two functions <span class="process-math">\(u(x)\text{,}\)</span> <span class="process-math">\(v(x)\in C[a,b]\text{,}\)</span> define the <dfn class="terminology">inner product</dfn> on <span class="process-math">\(C[a,b]\)</span> as</p>
<div class="displaymath process-math" data-contains-math-knowls="">
\begin{equation*}
&lt;u,v&gt;:=\int_a^b u(x) v(x)~\textrm{d}x.
\end{equation*}
</div>
<p class="continuation">In our case, <span class="process-math">\(a = -L\text{,}\)</span> <span class="process-math">\(b = L\)</span> (period <span class="process-math">\(2L\)</span>), i.e.,</p>
<div class="displaymath process-math" data-contains-math-knowls="">
\begin{equation*}
&lt;u,v&gt;:=\int_{-L}^{L} u(x) v(x)~\textrm{d}x.
\end{equation*}
</div>
<p class="continuation"><dfn class="terminology">Definition</dfn> The functions <span class="process-math">\(u(x)\)</span> and <span class="process-math">\(v(x)\)</span> are <dfn class="terminology">orthogonal</dfn> if <span class="process-math">\(&lt;u, v&gt; = 0\text{.}\)</span><dfn class="terminology">Claim 1:</dfn> The trig set is <em class="emphasis">mutually orthogonal</em>, i.e, any two distinct functions in the set are orthogonal to each other. This means</p>
<div class="displaymath process-math" data-contains-math-knowls="">
\begin{equation*}
\begin{aligned}
&lt;1,\sin \frac{m\pi x}{L}&gt;=0, \qquad &lt;1,\cos \frac{m\pi x}{L}&gt;=0,&amp; &amp; \forall m\\
&lt;\sin \frac{m\pi x}{L}, \sin \frac{n\pi x}{L}&gt;=0,  \quad &lt;\cos \frac{m\pi x}{L}, \cos \frac{n\pi x}{L}&gt;=0,&amp; &amp; \forall m\neq n\\
&lt;\sin \frac{m\pi x}{L}, \cos \frac{n\pi x}{L}&gt;=0,&amp; &amp; \forall m,n
\end{aligned}
\end{equation*}
</div>
<p class="continuation"><dfn class="terminology">Proof.</dfn> Check the inner product by direct integration:</p>
<div class="displaymath process-math" data-contains-math-knowls="">
\begin{equation*}
\int_{-L}^{L} \sin \frac{m\pi x}{L} \textrm{d}x =0, \quad \int_{-L}^{L} \cos \frac{m\pi x}{L} \textrm{d}x =0,~~~\forall m
\end{equation*}
</div>
<div class="displaymath process-math" data-contains-math-knowls="">
\begin{equation*}
\int_{-L}^{L} \sin \frac{m\pi x}{L}\sin \frac{n\pi x}{L} \textrm{d}x=\frac{1}{2}\int_{-L}^{L} \left[\cos\frac{(m-n)\pi x}{L} -\cos\frac{(m+n)\pi x}{L}\right] \textrm{d}x =0,~~\forall m\neq n.
\end{equation*}
</div>
<p class="continuation">All other identities can be proven in a similar way.<dfn class="terminology">Claim 2:</dfn> The inner product of any two equal functions in the trig set equals some constants, that is,</p>
<div class="displaymath process-math" data-contains-math-knowls="">
\begin{equation*}
&lt;1,1&gt;=2L,
\end{equation*}
</div>
<div class="displaymath process-math" data-contains-math-knowls="">
\begin{equation*}
&lt;\sin \frac{n\pi x}{L}, \sin \frac{n\pi x}{L}&gt;=L,  \qquad &lt;\cos \frac{n\pi x}{L}, \cos \frac{n\pi x}{L}&gt;=L, \quad \forall n
\end{equation*}
</div>
<p class="continuation"><dfn class="terminology">Proof.</dfn> Direct integration gives</p>
<div class="displaymath process-math" data-contains-math-knowls="">
\begin{equation*}
\int_{-L}^{L} \sin^2 \frac{n\pi x}{L} \textrm{d}x=\frac{1}{2}\int_{-L}^{L} \left[1 -\cos\frac{2n\pi x}{L}\right] \textrm{d}x =L.
\end{equation*}
</div>
<p class="continuation">We omit the proof for the other identities.</p>
<span class="incontext"><a href="sec7_3.html#p-340" class="internal">in-context</a></span>
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